Every finite Galois extension of $\mathbb{Q}_p$ has a solvable Galois group.

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I have found the statement as in the title of this post on wikipedia, however, there is no reference for its proof. How does one prove it?

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This follows from the theory of ramification groups (in the "lower numbering").

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Let $K/{\Bbb Q}_p$ be a Galois extension and $k$ the residue field of $K$. As Galois actions preserve integrality there is an induced action of $G$ (the Galois group) on ${\frak O}_K$ fixing $\Bbb Z_p$; it futhermore fixes powers of the maximal ideal of ${\frak O}_K$ and so induces an action on $k$ over ${\Bbb F}_p$, and more generally $G$ acts as automorphisms of ${\frak O}_K/{\frak m}^i$ for all $i\ge0$. We define $G_i:=\ker(G\to{\rm Aut}({\cal O}_K/{\frak m}^{i+1}))$ for $i\ge-1$ to be the ramification groups of $K/{\Bbb Q}_p$. We have $G_{-1}=G$, and $G_0$ is called the inertia group of the extension, while $G_1$ is called the wild inertia group. There is now a ramification filtration

$$G=G_{-1}\trianglerighteq G_0\trianglerighteq G_1\trianglerighteq G_2\trianglerighteq\cdots $$

Via $1$st iso thm there is a canonical isomorphism $G/G_0\cong{\rm Gal}(k/{\Bbb F}_p)\cong C_f$ where $f$ is the residue degree of $K/{\Bbb Q}_p$ (that is $f=\dim_{\,{\Bbb F}_p}k$). Then, if $\pi$ is a uniformizer (principal generator of ${\frak O}_K$'s max ideal ${\frak m}$), the map $\sigma\mapsto \sigma(\pi)/\pi$ is an injective group homomorphism $G_i/G_{i+1}\to U_i/U_{i+1}$ where $U_i:=1+{\frak m}^i$ are the higher unit groups (under multiplication). The quotient $U_0/U_1$ is clearly $k^\times\cong C_{p^f-1}$, and $U_i/U_{i+1}\hookrightarrow {\frak m}^i/{\frak m}^{i+1}\cong k^+\cong C_p^f$ hence is elementary abelian.

Thus $G_1\hookrightarrow G_0\hookrightarrow G$ where $G_1$ is a $p$-group (hence solvable), $G_0/G_1$ is cyclic of order dividing $|k^\times|=p^f-1$ and $G/G_0$ is cyclic of order $f$. Therefore, $G$ is a solvable group.

For more detail and some proofs, I like Boston's The Proof of Fermat's Last Theorem's Chapter 3, from Definition 3.6 up to Corollary 3.9 (which is the desired claim). (The link is an official pdf.)